Find the LCM 2x^2-2 and 4x-4

$2 x^{2} - 2$ and $4 x - 4$

Factor $2 x^{2} - 2$ .

Factor $2$ out of $2 x^{2} - 2$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) - 2$

Factor $2$ out of $- 2$ .

$2 (x^{2}) + 2 (- 1)$

Factor $2$ out of $2 (x^{2}) + 2 (- 1)$ .

$2 (x^{2} - 1)$

Rewrite $1$ as $1^{2}$ .

$2 (x^{2} - 1^{2})$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$2 ((x + 1) (x - 1))$

Remove unnecessary parentheses.

$2 (x + 1) (x - 1)$

Factor $4$ out of $4 x - 4$ .

Factor $4$ out of $4 x$ .

$4 (x) - 4$

Factor $4$ out of $- 4$ .

$4 (x) + 4 (- 1)$

Factor $4$ out of $4 (x) + 4 (- 1)$ .

$4 (x - 1)$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Multiply $2$ by $2$ .

$4$

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

The LCM of $x + 1, x - 1, x - 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 1) (x - 1)$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$4 (x + 1) (x - 1)$